Didactica De Las Maemáticas

An Introduction to p-adic Numbers and p-adic Analysis Andrew Baker
[10/02/2010]

Department of Mathematics, University of Glasgow, Glasgow G12 8QW, Scotland. E-mail address: a.baker@maths.gla.ac.uk URL: http://www.maths.gla.ac.uk/∼ajb

Contents
Introduction Chapter 1. Chapter 2. Chapter 3. Chapter 4. Congruences and modular equations The p-adic norm and the p-adic numbers Some elementaryp-adic analysis The topology of Qp 1 3 15 29 33 47 53 55 55 56 56 59 60

Chapter 5. p-adic algebraic number theory Bibliography Problems Problem Set 1 Problem Set 2 Problem Set 3 Problem Set 4 Problem Set 5

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Introduction
These notes were written for a final year undergraduate course taught at Manchester University in 1988/9 and also taught in later years by Dr M. McCrudden. I rewrotethem in 2000 to make them available to interested graduate students. The approach taken is very down to earth and makes few assumptions beyond standard undergraduate analysis and algebra. Because of this the course was as self contained as possible, covering basic number theory and analytic ideas which would probably be familiar to more advanced readers. The problem sets are based on those forproduced for the course. I would like to thank Javier Diaz-Vargas, Juan L´pez and Jeremy Scofield for pointing out o numerous errors. Updates of 24/03/2009: Slight rearrangement of the problems, and added material at end of Chapter 4 on p-adic exponential, logarithm and Artin-Hasse series.

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CHAPTER 1

Congruences and modular equations
Let n ∈ Z (we will usually have n > 0). We define thebinary relation ≡ by
n

Definition 1.1. If x, y ∈ Z, then x ≡ y if and only if n | (x − y). This is often also written x ≡ y (mod n) or x ≡ y (n).
n n

Notice that when n = 0, x ≡ y if and only if x = y, so in that case ≡ is really just equality.
0

Proposition 1.2. The relation ≡ is an equivalence relation on Z.
n

Proof. Let x, y, z ∈ Z. Clearly ≡ is reflexive since n | (x − x) = 0. It issymmetric since if n | (x − y) then x − y = kn for some k ∈ Z, hence y − x = (−k)n and so n | (y − x). For transitivity, suppose that n | (x − y) and n | (y − z); then since x − z = (x − y) + (y − z) we have n | (x − z). If n > 0, we denote the equivalence class of x ∈ Z by [x]n or just [x] if n is understood; it is also common to use x for this if the value of n is clear from the context. Fromthe definition, [x]n = {y ∈ Z : y ≡ x} = {y ∈ Z : y = x + kn for some k ∈ Z},
n n

and there are exactly |n| such residue classes, namely [0]n , [1]n , . . . , [n − 1]n . Of course we can replace these representatives by any others as required. Definition 1.3. The set of all residue classes of Z modulo n is Z/n = {[x]n : x = 0, 1, . . . , n − 1}. If n = 0 we interpret Z/0 as Z. Consider thefunction πn : Z −→ Z/n; This is onto and also satisfies
−1 πn (α) = {x ∈ Z : x ∈ α}.

πn (x) = [x]n .

We can define addition + and multiplication × on Z/n by the formulæ
n n

[x]n +[y]n = [x + y]n ,
n

[x]n ×[y]n = [xy]n ,
n

which are easily seen to be well defined, i.e., they do not depend on the choice of representatives x, y. The straightforward proof of our next result is left to thereader.
3

Proposition 1.4. The set Z/n with the operations + and × is a commutative ring and the function πn : Z −→ Z/n is a ring homomorphism which is surjective (onto) and has kernel ker πn = [0]n = {x ∈ Z : x ≡ 0}.
n n n

Now let us consider the structure of the ring Z/n. The zero is 0 = [0]n and the unity is 1 = [1]n . We may also ask about units and zero divisors. In the following, letR be a commutative ring with unity 1 (which we assume is not equal to 0). Definition 1.5. An element u ∈ R is a unit if there exists a v ∈ R satisfying uv = vu = 1. Such a v is necessarily unique and is called the inverse of u and is usually denoted u−1 . Definition 1.6. z ∈ R is a zero divisor if there exists at least one w ∈ R with w ̸= 0 and zw = 0. There may be lots of such w for each zero...